A point in rectangular coordinates is given.Convert the point to polar coordinates,r 0. $( - 13 , - 13 )$
A)Polar coordinates: $\left( \sqrt { 338 } , \frac { 5 \pi } { 4 } \right)$
B)Polar coordinates: $\left( - \sqrt { 338 } , \frac { 5 \pi } { 4 } \right)$
C)Polar coordinates: $\left( - \sqrt { 13 } , \frac { \pi } { 4 } \right)$
D)Polar coordinates: $\left( \sqrt { 13 } , \frac { \pi } { 4 } \right)$
E)Polar coordinates: $\left( \sqrt { 338 } , - \frac { \pi } { 4 } \right)$

Question

A point in rectangular coordinates is given.Convert the point to polar coordinates,r > 0.​ $( - 13 , - 13 )$ ​
A)Polar coordinates: $\left( \sqrt { 338 } , \frac { 5 \pi } { 4 } \right)$
B)Polar coordinates: $\left( - \sqrt { 338 } , \frac { 5 \pi } { 4 } \right)$
C)Polar coordinates: $\left( - \sqrt { 13 } , \frac { \pi } { 4 } \right)$
D)Polar coordinates: $\left( \sqrt { 13 } , \frac { \pi } { 4 } \right)$
E)Polar coordinates: $\left( \sqrt { 338 } , - \frac { \pi } { 4 } \right)$

Quizplus · Accepted Answer

The magnitude $r$ of the polar coordinates is calculated using the formula $r = \sqrt{x^2 + y^2}$, which gives $\sqrt{(-13)^2 + (-13)^2} = \sqrt{338}$. The angle $\theta$ in radians for a point in the third quadrant (both $x$ and $y$ negative) is $\frac{5\pi}{4}$.

A Point in Rectangular Coordinates Is Given (−13,−13)( - 13 , - 13 )(−13,−13)

A Point in Rectangular Coordinates Is Given $( - 13 , - 13 )$